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If z = 3, what is 5 x (6 – z)?
If z = 3, what is 5 x (6 – z)?
How Subtract 7 from z?
z - 7Example: if z = 20z - 7 = 20 - 7 = 13
If Negative 9 times z plus 4 equals Negative 5 what does z equal?
Z= -1
Write a variable expression for this amount... 3 times the sum of a number z and 15?
3(z+15)
What does 3(z + 5)?
8
Find the value of z that makes the equation true. 5 x z plus 3 23?
To find the value of z that makes the equation true, we need to solve for z in the equation 5z + 3 = 23. First, subtract 3 from both sides to isolate the term with z: 5z = 20. Then, divide both sides by 5 to solve for z: z = 4. Therefore, the value of z that makes the equation true is z = 4.
If z = 3, what is 5 x (6 – z)?
If z = 3, what is 5 x (6 – z)?
How do you write an algebreic expression with 5 less than 3 times a number z?
3z-5
Find an algebraic expression for the quotient of 3 and sum of z and 5?
The algebraic expression for the quotient of 3 and the sum of z and 5 is 3 / (z + 5). In this expression, the numerator is 3, and the denominator is the sum of z and 5. This expression represents dividing 3 by the sum of z and 5.
How do you you subtract a mixed number?
23/8990 += z h20 x + 2 times z square denominator s=+ hoxz/ square root so i'ts 55507+h20xz rooT Z
Can you understand my proof about Fermat?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
What are the efforts of Africa to solve the global news flow disequilibrium?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
How is newest proof Fermat?
Pierre De Fermat 's last Theorem. The conditions: x,y,z,n are the integers and >0. n>2. Proof: z^n=/x^n+y^n. We have; z^3=[z(z+1)/2]^2-[(z-1)z/2]^2 Example; 5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125 And z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2 Example; 5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189 And z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2 Example 5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216 And z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2 Example 5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224 General: z^3+(z-1)^3+....+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2 We have; z^3=z^3+(z-m-1)^3 - (z-m-1)^3. Because: z^3+(z-m-1)^3=[z^3+(z-1)^3+....+(z-m-1)^3] - [(z-1)^3+....+(z-m)^3] So z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3. Similar: z^3=z^3+(z-m-2)^3 - (z-m-2)^3. So z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3. .... .... Suppose: z^n=x^n+y^n So z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3. So z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 - [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 - (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 - [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 - (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 - [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 - (y-m-1)^3} Similar: z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 - [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 - (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 - [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 - (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 - [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 - (y-m-2)^3. .... .... Because it is codified . So Impossible all are the integers. So: z^n=/x^n+y^n. ISHTAR.
What times what minus what would equal to 13?
To solve the equation "x times y minus z equals 13," we need to set up the equation as follows: xy - z = 13. Since we are looking for three unknowns, there are multiple solutions to this equation. One possible solution could be x = 5, y = 3, and z = 2, as 5 times 3 minus 2 equals 13.
How Subtract 7 from z?
z - 7Example: if z = 20z - 7 = 20 - 7 = 13
Can anyone code this in C plus plus output 1 5 2 4 3 3 2 4 1 5?
int i, x, y, z; x=1; y=5; z=1; for (i=0; i<5; ++i) { printf ("%d %d ", x, y); if (x==3) z=z*(-1); x=x+z; y=y-z; } Output: 1 5 2 4 3 3 4 2 5 1
If Negative 9 times z plus 4 equals Negative 5 what does z equal?
Z= -1
